How do we know that $\sum_{k=0}^{\infty}\frac{x^k}{k!}=e^x$?

Question

I've been taught that the definition of the exponential function is the following power series: $$\sum_{k=0}^{\infty}\frac{x^k}{k!}$$ Here's my question: how do we know that this series is equal to $e^x$? That is to say, how do we know that the function defined by this power series can be expressed in the form $c^x$ where $c$ is a constant and that $c=e$? If we look at another power series, say, $\sum_{k=0}^{\infty}(-1)^k\frac{x^{2k}}{2k!}$, there is obviously no constant $c$ for which $c^x=\sum_{k=0}^{\infty}(-1)^k\frac{x^{2k}}{2k!}$, so what tells us that there exists such a constant for the first series?

Answer 1

It is only after some manipulation that it becomes something familiar. Your other example of course is $\cos x$.
A common approach is to find what differential equation it satisfies. Let $y(x) = \sum_{k=0}^{\infty}\frac{x^k}{k!}$.
Differentiate it term by term, and you find $\frac{dy}{dx}=y(x)$.
That is a differential equation. It is separable, and you get $x=c+\log y$. You can tell $c=0$ because $y=1$ when $x=0$. Lastly, the inverse function of $\log x$ is $e^x$.
Your other series is a solution to $\frac{d^2y}{dx^2}+y=0$, and so are $\cos x$ and $\sin x$, so it must be one of those, or a combination of those.